Nilai dari integral batas atas pi/4 batas bawah 0 1

Berikut ini adalah pertanyaan dari JefferySantosa pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

Nilai dari integral batas atas pi/4 batas bawah 0 1 / 9cos^2(x) - sin^2(x)

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

$\large\text{$\begin{aligned}&\int_{0}^{\pi/4}\frac{1}{9\cos^{2}x-\sin^{2}x}\,dx\\&=\ \boxed{\,\bf\frac{ln\,(2)}{6}\,}\end{aligned}$}$
 

Pembahasan

Integral Tentu

$\begin{aligned}&\int_{0}^{\pi/4}\frac{1}{9\cos^{2}x-\sin^{2}x}\,dx\\{=\ }&\int_{0}^{\pi/4}\frac{1}{(3\cos x+\sin x)(3\cos x-\sin x)}\,dx\\&\ \left[\ \begin{aligned}\sin x=\frac{\tan x}{\sec x},\ \cos x=\frac{1}{\sec x}\end{aligned}\right.\\{=\ }&\int_{0}^{\pi/4}\frac{1}{\left(\dfrac{3+\tan x}{\sec x}\right)\left(\dfrac{3-\tan x}{\sec x}\right)}\,dx\\{=\ }&\int_{0}^{\pi/4}\frac{1}{(3+\tan x)(3-\tan x)}\,\sec^2x\,dx\end{aligned}$
$\begin{aligned}&\ \left[\ \begin{aligned}&{\sf Ambil\ }u=\tan x\\&\Rightarrow du=\sec^2x\,dx\\\end{aligned}\right.\\{=\ }&\int_{u=\tan0}^{u=\tan\pi/4}\frac{1}{(3+u)(3-u)}\,du\\{=\ }&{-}\int_{0}^{1}\frac{1}{(u+3)(u-3)}\,du\quad...(i)\\\end{aligned}$

Sampai di sini kita bisa menggunakan pecahan parsial atau cara lain.

Cara 1: Dengan pecahan parsial

$\begin{aligned}&&\!\!\!\!\!\!\!\!\!\!\frac{1}{(u+3)(u-3)}&=\frac{A}{u+3}+\frac{B}{u-3}\\&&&=\frac{A(u-3)+B(u+3)}{(u+3)(u-3)}\\&\Rightarrow&1&=A(u-3)+B(u+3)\\&\Rightarrow&u=-3&\implies A=-\frac{1}{6}\\&&u=3&\implies B=\frac{1}{6}\end{aligned}$

Substitusi ke dalam $(i)$ :

$\begin{aligned}&{-}\int_{0}^{1}\frac{1}{(u+3)(u-3)}\,du\\{=\ }&{-}\int_{0}^{1}\left(-\frac{1}{6(u+3)}+\frac{1}{6(u-3)}\right)du\\{=\ }&\frac{1}{6}\cdot\int_{0}^{1}\left(\frac{1}{u+3}-\frac{1}{u-3}\right)du\\&\ \left[\ \begin{aligned}&{\sf Ambil\ }v=u+3\\&\Rightarrow dv=du\\&{\sf Ambil\ }w=u-3\\&\Rightarrow dw=du\\\end{aligned}\right.\\{=\ }&\frac{1}{6}\cdot\int_{u=0}^{u=1}\left(\frac{1}{v}\,dv-\frac{1}{w}\,dw\right)\\{=\ }&\frac{1}{6}\Big[\ln|v|-\ln|w|\Big]_{u=0}^{u=1}\end{aligned}$
$\begin{aligned}{=\ }&\frac{1}{6}\Big[\ln|u+3|-\ln|u-3|\Big]_{0}^{1}\\{=\ }&\frac{1}{6}\left(\ln|4|-\ln|{-}2|-\cancel{\ln|3|}+\cancel{\ln|{-}3|}\right)\\{=\ }&\frac{1}{6}\left(\ln(4)-\ln(2)\right)\\{=\ }&\frac{1}{6}\cdot\ln\left(\frac{4}{2}\right)\\{=\ }&\boxed{\,\bf\frac{ln\,(2)}{6}\,}\end{aligned}$

$\blacksquare$

Cara 2

Dari $(i)$ :

$\begin{aligned}&{-}\int_{0}^{1}\frac{1}{(u+3)(u-3)}\,du\\&\ \left[\ \begin{aligned}&{\sf Ambil\ }v=u+3\\&\Rightarrow u-3=v-6\\&\Rightarrow dv=du\\\end{aligned}\right.\\{=\ }&{-}\int_{u=0}^{u=1}\frac{1}{v(v-6)}\,dv\\{=\ }&{-}\int_{u=0}^{u=1}\frac{1}{v^2\left(1-\dfrac{6}{v}\right)}\,dv\\{=\ }&{-}\int_{u=0}^{u=1}\frac{1}{\left(1-\dfrac{6}{v}\right)}\cdot\frac{1}{v^2}\,dv\end{aligned}$
$\begin{aligned}&\ \left[\ \begin{aligned}&{\sf Ambil\ }w=1-\frac{6}{v}\\&\Rightarrow dw=\frac{6}{v^2}\,dv=6\cdot\frac{1}{v^2}\,dv\\&\Rightarrow \frac{1}{6}\,dw=\frac{1}{v^2}\,dv\end{aligned}\right.\\{=\ }&{-}\int_{u=0}^{u=1}\frac{1}{w}\cdot\frac{1}{6}\,dw\\{=\ }&{-}\frac{1}{6}\Big[\ln|w|\Big]_{u=0}^{u=1}\\{=\ }&{-}\frac{1}{6}\left[\ln\left|1-\frac{6}{v}\right|\right]_{u=0}^{u=1}\\{=\ }&{-}\frac{1}{6}\left[\ln\left|\frac{v-6}{v}\right|\right]_{u=0}^{u=1}\end{aligned}$
$\begin{aligned}{=\ }&{-}\frac{1}{6}\left[\ln\left|\frac{u-3}{u+3}\right|\right]_{0}^{1}\\{=\ }&{-}\frac{1}{6}\left(\ln\left|\frac{-2}{4}\right|-\ln\left|\frac{-3}{3}\right|\right)\\{=\ }&{-}\frac{1}{6}\left(\ln\left(\frac{1}{2}\right)-\ln(1)\right)\\{=\ }&{-}\frac{1}{6}\cdot\ln\left(\frac{1/2}{1}\right)={-}\frac{1}{6}\cdot\ln\left(\frac{1}{2}\right)\\{=\ }&{-}\frac{1}{6}\left(-\ln(2)\right)\\{=\ }&\boxed{\,\bf\frac{ln\,(2)}{6}\,}\end{aligned}$